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The exercise is the exercise 6.6 from Evans' PDE.

Suppose $U$ is connected, and $\partial U$ consists of two disjoint, closed sets $\Gamma_1$ and $\Gamma_2$. Define what it means for $u$ to be a weak solution of Poisson equation with mixed Dirichlet-Neumann boundary conditions: $$ \begin{cases} -\Delta u = f \ \ \ \text{in $U$} \\ u = 0 \ \ \ \text{on $\Gamma_1$} \\ \frac{\partial u}{\partial \nu} = 0 \ \ \text{on $\Gamma_2$}. \end{cases} $$

My attempts: Let $u \in C^{\infty}(U)$ be a solution to the above problem. Then for $v \in H^1(U)$, integration by parts yields \begin{align} (f,v) = -\int_U (\Delta u) v & = \int_U Du \cdot Dv -\int_{\partial U} \frac{\partial u}{\partial \nu} v \\ & = \int_U Du\cdot Dv - \int_{\Gamma_1} \frac{\partial u}{\partial \nu} v \end{align}

I wish to conclude $\int_{\Gamma_1} \frac{\partial u}{\partial \nu} v = 0$, but I don't know how to do that. Could anyone give me some hint?

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You can pick the space of test functions $v$ to be the space $$H^1_{\Gamma_1}(U) = \{v \in H^1(U)\colon v = 0 \text{ on } \Gamma_1 \}, $$ which is a Hilbert space with respect to the inner product in $H^1$. Actually, you can verify that in $H^1_{\Gamma_1}$ the norm $$ \|v\|_{H^1_{\Gamma_1}(U)} = \|Dv\|_{L^2(U)}$$ is equivalent to the $H^1$ norm (by Poincaré inequality). Then the second integral equals $0$ since $v \in H^1_{\Gamma_1}(U).$

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  • $\begingroup$ Thank you! May I ask that: what's the explicit definition of your "scalar product in $H_1$"? According to the context, it seems that you're not referring to the $H^1$ inner product. But I was thinking if we could use $H^1$ inner product directly. It suffices to check the completeness. Since $H^1$ is already complete, suffices to check closeness. If $u_n \in H^1_{\Gamma_1} \to u$ in $H^1$, then by trace theorem, $u_n \to u$ in $L^2(\Gamma_1)$. Then certainly there has a subsequence $u_n \to u$ a.e. in $\Gamma_1$. That is, $u = 0$ at $\Gamma_1$, saying that $u \in H^1_{\Gamma_1}$. Am I correct? $\endgroup$
    – mathdoge
    Jul 16, 2020 at 14:12
  • $\begingroup$ Yes, you are. With "scalar product" I meant "inner product". I edit my answer. $\endgroup$
    – G. Gare
    Jul 16, 2020 at 14:14

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